852 Appendix II.
any primitive curve, Fig. 815 : erect several vertical lines,
unequally spaced, there being more where the curve varies greatly,
also further mid verticals n', 22', &c. Take any pole o, at
a known distance p from a vertical XY : project points i', 2',
3', &c., horizontally to XY, and join each projection to o.
Next draw A# II 01", a6\\o2", bc\\Q$", &c., till the whole
sum curve A to k be formed, whose ordinate at any point will
shew area under primitive curve up to that point, measuring from
A ; while DJ will represent the whole area A B c D.
To Draw the Slope Curve, the reverse process must be per-
formed, and thus the primitive curve in Fig. 815 will shew by its
ordinates the slope of the sum curve. For example, suppose we
want the slope between c d, we draw 04" || c d, and produce 4"
to 4', giving a point in what will be the slope curve.
For proof draw cc± horizontally. Then
slope vied = dc± = ill = vertical of^lope curve
cc^ p p
If we make /i = i, the steepness of any primitive curve will be
measured by the slope curve ordinate. Again, by cross multipli-
But cc± x 44' is the area between two verticals of primitive curve.
Hence if/=i, dc^Qi rise of sum curve will shew increase of
primitive curve area, and ordinate of sum curve at any point will
shew area of primitive curve to that point. Also the value/ may
be so arranged that the new curves may be read off directly to any
Bending Moment and Vertical Shear by Graphic
Summation. — The shear on any beam section is due to total
load on one side of section less the support reaction (if any) on
that side. The beam A c, Fig. 816, is stressed by a load whose
intensity per foot run is shewn by curve ABC, drawn to a scale of
100 tons per inch. Base scale being i foot per inch or ^, let
, f in. shew 100 tons on shear scale. Supposing an increase of 100
tons on the shear curve, over a base of i foot; d = 100 tons, £ =
- in,, and y (load) = 100 tons = i in.